Telescoping Sum of expectations: limsup exists but limes not necessarily Let $X_t$ for $t \in \{0, 1, \dotsc, \}$ be a sequence of  non-negative integer-valued random variables. Suppose that $$\mathbb{E}[X_t - X_{t+1} \mid X_t>0 ] \leq c \quad \text{  for some constant }  c \text{, for all } t \geq 0$$
and for all possible outcomes $\omega \in \Omega$.
Moreover assume that 
$$\liminf_{t \rightarrow \infty} \mathbb{E}[X_t] = 0.$$ 
Can we show that 
$$\sum_{t=0}^{\infty} \left(\mathbb{E}[X_t]- \mathbb{E}[X_{t+1}]\right)\geq \mathbb{E}[X_0]?$$
Concrete Example: Random Declines. 
Let $X_0=n$ for $n \rightarrow \infty$. Hence we have a deterministic starting value. Then choose $X_{t+1}$ uniformly at random from the set $\{0, \dotsc, \lfloor e X_t \rfloor\}$ for all $t \geq 0$. Can we say that $\sum_{t=0}^{\infty} (\mathbb{E}[X_t]-\mathbb{E}[X_{t+1}])\geq \mathbb{E}[X_0]= n$?
 A: Think about the sequence of independent random variables $(X_t)_{t \in \mathbb{N}}$ where
\begin{align}
X_t & \sim Unif((1,2)) &~\text{if}~& t=2k \\
X_t & \sim Unif((0,\frac{1}{t})) &~\text{if}~& t=2k + 1 
\end{align}
for any $k \in \mathbb{N}$. Then your sum does not converge. 
If you want to prove that sum 
$$\sum_{n=1}^{\infty}a_n$$ 
converge (so you can define something using it) you have to know (prove) that
$$\lim_{n \to \infty} a_n = 0.$$
Otherwise the sum is not defined. 
Edit:
Since OP put another assumption I have to create another counterexample: think about the sequence of independent random variables $(X_t)_{t \in \mathbb{N}}$  such that 
\begin{align}
X_t & \sim Y &~\text{if}~& t=2k \\
X_t & \sim \frac{1}{t}.Y &~\text{if}~& t=2k + 1 
\end{align}
where $Y$ has Bernouli distribution with $p=0.5$ Again your sum does not converge. 
Existence of limit of $a_n$:
If 
$$\sum_{n=1}^{\infty}a_n = \lim_{k \to \infty}  \sum_{n=1}^{k}a_n $$ 
exists, than for all $m >1$ 
$$|a_m| \leq \left| \sum_{n=1}^{m}a_n - \sum_{n=1}^{m-1}a_n \right|. $$
Since you assume that the infinite sum converge, the right hand side of the previous inequality goes to $0$ as $m \to \infty$. So the 
$$\lim_{m \to \infty} |a_m| = 0.$$
Edit 2:
Counterexample no. 3: define sequence of independent random variables $(X_t)_{t \in \mathbb{N}}$, such that every $X_t$ has Bernoulli distribution with parameter $p_t$. Define these parameters as follows
\begin{align}
p_t &=\frac{1}{2} &~\text{if}~& t=2k, \\
p_t &= \frac{1}{t} &~\text{if}~& t=2k + 1. 
\end{align}
Again the desired sum does not converge. 
